poj 2247

枚举，貌似数据量挺大的，其实不大，就看最深层的tot加了多少次，就可估计出真正的规模

//枚举
#include <iostream>
#include <stdio.h>
#include <algorithm>
#define ll long long
using namespace std;
const int maxn=2000000000;
int num[10000];
int main()
{	
	ll i1,i2,i3,i4;//注意long long 因为虽然按理说是不会爆int，相乘这个过程可能会溢出
	int tot=0;
	for(i1=1;i1<=maxn;i1*=2)
	{
		for(i2=1;i1*i2<=maxn;i2*=3)
		{
			for(i3=1;i1*i2*i3<=maxn;i3*=5)
			{
				for(i4=1;i1*i2*i3*i4<=maxn;i4*=7)
					num[tot++]=i1*i2*i3*i4;
			}
		}
	}
	sort(num,num+tot);
	int n;
	while(scanf("%d",&n)&&n)
	{
		if(n%100==11) printf("The %dth humble number is %d.\n",n,num[n-1]);
		else if(n%100==12) printf("The %dth humble number is %d.\n",n,num[n-1]);
		else if(n%100==13) printf("The %dth humble number is %d.\n",n,num[n-1]);
		else if(n%10==1) printf("The %dst humble number is %d.\n",n,num[n-1]);
		else if(n%10==2) printf("The %dnd humble number is %d.\n",n,num[n-1]);
		else if(n%10==3) printf("The %drd humble number is %d.\n",n,num[n-1]);
		else printf("The %dth humble number is %d.\n",n,num[n-1]);
	}
	return 0;
}